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2 years ago

Estimate the number 78% of 112

Mathematics

2 answers:

andreyandreev [35.5K]2 years ago

4 0

I believe it is 87.36. You would take 78% and muiltply it by 112. I think. Hope I helped.

balu736 [363]2 years ago

3 0

To find 78% of 112

find 10% which is 11.2 (divide by 10)
find 70% by multiplying 11.2 by 7 which is 78.4
find 1% by dividing it by 100 which would give you 1.12
find 8% by multiplying 1.12 by 8 which would be 8.96
78% is 78.4 + 8.96 which is 87.36

ANSWER- 87.36

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Evaluate 5 - 44 * (-0.75) - 18 / 2/3 * 0.8 + (-4/5)

jenyasd209 [6]

Answer:

15.6

step by step explanation:

5-44• (-0.75) - 18 ÷ 2/3 • 0.8 + (-4/5)

= 5 + 33 - 27*0.8 - 0.8

= 5 + 33 -21.6 - 0.8

= 15.6.

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2 years ago

Two competing gyms each offer childcare while parents work out gym a charged $9.00 per hour of childcare. Gym b charges $0.75 pe

andriy [413]

Answer:

D.)Gym B and Gym A charge the same hourly rate for childcare.

Step-by-step explanation:

You need to find out how much Gym B charges for a full hour. There are 60 minutes in an hour. Divide this by 5 to determine how many 5 minute increments are in an hour: 12. A parent paying for 1 hour of child care will need to make 12 payments of .75. 12 times .75 equals $9.00. This means that both gyms charge the same amount per hour.

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2 years ago

Prove the divisibility of the following numbers:

ratelena [41]

Make use of prime factorizations:

$16^5+2^{15}=(2^4)^5+2^{15}=2^{20}+2^{15}$

Both terms have a common factor of $2^{15}$ :

$16^5+2^{15}=2^{15}\left(2^5+1\right)=2^{15}\cdot33$

- - -

The second one is not true! We can write

$15^7+5^{13}=(3\cdot5)^7+5^{13}=3^7\cdot5^7+5^{13}$

Both terms have a common factor of $5^7$ :

$15^7+5^{13}=5^7\left(3^7+5^6\right)$

Since $30=5\cdot6$ , and $5\mid5^7$ , we'd still have to show that $5^6(3^7+5^6)$ is a multiple of 6. This is impossible, because $6=3\cdot2$ and there is no multiple of 2 that can be factored out.

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2 years ago

Read 2 more answers

Can someone tell me how

gulaghasi [49]

the way I get the subsequent term, nevermind the exponents, the exponents part is easy, since one is decreasing and another is increasing, but the coefficient, to get it, what I usually do is.

multiply the current coefficient by the exponent of the first-term, and divide that by the exponent of the second-term + 1.

so if my current expanded term is say 7a³b⁴, to get the next coefficient, what I do is (7*3)/5 <----- notice, current coefficient times 3 divided by 4+1.

anyhow, with that out of the way, lemme proceed in this one.

$\bf ~~~~~~~~\textit{binomial theorem expansion} \\\\ \qquad \qquad (1+ax)^n\implies \begin{array}{llll} term&coefficient&value\\ \cline{1-3}&\\ 1&+1&(1)^n(ax)^0\\\\ 2&+\frac{(1)(n)}{1}\to n&(1)^{n-1}(ax)^1\\\\ 3&+\frac{n\cdot (n-1)}{2}&(1)^{n-2}(ax)^2 \end{array}$

so, following that to get the next coefficient, we get those equivalents as you see there for the 2nd and 3rd terms.

so then, we know that the expanded 2nd term is 24x therefore

$\bf n(1)^{n-1}(ax)1 = 24x\implies n(1)(ax)=24x\implies nax=24x\implies n=\cfrac{24}{a}$

we also know that the expanded 3rd term is 240x², therefore we can say that

$\bf \cfrac{n(n-1)}{2}~~(1)^{n-2}(ax)^2 = 240x^2\implies \cfrac{n(n-1)}{2}(1)(a^2x^2) = 240x^2 \\\\\\ \cfrac{(n^2-n)(a^2x^2)}{2}=240x^2\implies \cfrac{(n^2-n)(a^2)}{2}=\cfrac{240x^2}{x^2}\implies \cfrac{a^2n^2-a^2n}{2}=240 \\\\\\ a^2n^2-a^2n=480$

but but but, we know what "n" equals to, recall above, so let's do some quick substitution

$\bf a^2n^2-a^2n=480\qquad \boxed{n=\cfrac{24}{a}}\qquad a^2\left( \cfrac{24}{a} \right)^2-a^2\left( \cfrac{24}{a} \right)=480 \\\\\\ a^2\cdot \cfrac{24^2}{a^2}-24a=480\implies 24^2-24a=480\implies 576-24a=480 \\\\\\ -24a=-96\implies a=\cfrac{-96}{-24}\implies \blacktriangleright a = 4\blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ n=\cfrac{24}{a}\implies n=\cfrac{24}{4}\implies \blacktriangleright n=6 \blacktriangleleft$

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2 years ago

Explain how to find each root, with its multiplicity, for the polynomial y=x^5-27x^2.

Nuetrik [128]

The roots of $y = x^{5}-27\cdot x^{2}$ are 0 (multiplicity 2), -3 (multiplicity 1), $-\frac{3}{2}+i\,\frac{3\sqrt{3}}{2}$ (multiplicity 1) and $-\frac{3}{2}-i\,\frac{3\sqrt{3}}{2}$ (multiplicity 1).

The characteristics of the polynomial can be derived from algebraic techniques. Roots and multiplicity can be found by <em>factoring</em> the polynomial. The multiplicity is represented by a power binomial of the form:

$(x-r_{i})^{m},\,m \le n$ (1)

Where $n$ is the grade of the polynomial.

Now we proceed to factor the formula:

$y = x^{5}-27\cdot x^{2}$

$y = x^{2}\cdot (x^{3}-27)$

$y = x^{2}\cdot (x^{2}+3\cdot x +9)\cdot (x-3)$

Please notice that $x^{2}+3\cdot x + 9$ have two complex roots.

$y = x^{2}\cdot \left(x+\frac{3}{2}-i\,\frac{3\sqrt{3}}{2} \right)\cdot \left(x+\frac{3}{2}+i\,\frac{3\sqrt{3}}{3} \right)\cdot (x-3)$

The roots of $y = x^{5}-27\cdot x^{2}$ are 0 (multiplicity 2), -3 (multiplicity 1), $-\frac{3}{2}+i\,\frac{3\sqrt{3}}{2}$ (multiplicity 1) and $-\frac{3}{2}-i\,\frac{3\sqrt{3}}{2}$ (multiplicity 1).

We kindly invite to see this question on polynomials: brainly.com/question/1218505

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2 years ago